The net dipole moment of the molecule Atenolol has amide functional group that makes it the most polar out of three drugs. Because of the greater electronegativity of oxygen‚ the carbonyl (C=O) is a stronger dipole than the N–C dipole. The presence of a C=O dipole and‚ to a lesser extent N–C dipole‚ allows amides to act as H-bond acceptors. The presence of N–H dipoles allows amides to function as H-bond
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Biology Honors Final Exam Review Study Guide A. Intro Unit * Characteristics of life (DR.CHARGE) * DNA * Directs protein synthesis * Reproduce * Divides in mitosis (asexual reproduction) * Cells * Homeostasis * The process by which cells maintain constant internal condition (water‚ temperature) * Adaptation (Evolution) * Respond to stimulus * Growth (Development) * Energy * Heterotroph: an organism that cannot synthesize its own food
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9/22/2011 Basic Chemistry Why are we studying chemistry? • Biology has chemistry at its foundation Elements • All matter composed of basic substances called elements • Living organisms require about twenty of the known elements Together‚ carbon‚ hydrogen‚ nitrogen and oxygen make up 95% of the elements found in the body. Atoms • All elements made of tiny particles called atoms • Made up of: – Nucleus • contains protons (+ charge) and neutrons (no charge) – Outside
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valence electrons by two atoms. Molecule - two or more atoms held together by covalent bond. Single Bond - a pair of shared electrons Double bond - two pairs of valence electrons are shared. Valence - the bonding capacity of an element. Electronegativity 0 the attraction of a
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more electrons from a metal onto a non-metal. Characteristics of both bonds: * Occur between 2 atoms * Composed of 2 electrons * Have both ionic and covalent characteristics * Together = 100% * Both bonds are measured on an electronegativity scale * Both contain a nonmetal * Chemical bonds * Are determined by using the “magic number” (1.67) * Have bond angle and bond axis Characteristics of metallic bonds: * In metals (d-block electrons only..because of shape and
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Discussion: In this experiment a ketone‚ 9-fluorenone is reduced to and alcohol. The are two possible ways by which this reduction can occur. One is by a catalytic hydrogenation‚ this uses a catalyst such as palladium or nickel‚ hydrogen gas‚ and heat/pressure. This can reduced an alkane to alkene. This catalytic process is preferred in industrial practices because the cost is low in the long run and more importantly there is little to no waste expense. However‚ hydrogen gas is dangerous due to
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Element name-Helium Element Symbol-He Atomic number-2 Group name or number- Part of the noble gases in group 18 Atomic mass (to nearest thousandth)-4.003 Number of protons-2 Number of neutrons-2 Number of electrons-2 Number of valence electrons-2 List of isotopes-He3 and He4 Electron Configuration-1s2 Discovered By-Pierre Janssen Year discovered-1868 Circumstances of discovery- Janssen discovered helium in 1868 when he was looking in a telescope when he found the yellow spectrum lines of helium
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glucose is: 180 g/mol How many grams of glucose should you weigh out and add to 500 ml to make a 25 mM solution 2.25 g Your final solution has a [H+] = 3.98x10^-8 what is the pH of the solution? pH= 7.4 Carbon and nitrogen have dissimilar electronegativities. When these atoms combine together what type of bond forms between them Polar covalent What type of bond is formed from the transfer of electrons between elements Ionic How many chloride atoms
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electronegative then oxygen. We know that acidity increases from left to right in the periodic table‚ which would mean that as we move right to left (Nitrogen is to the left of oxygen)‚ compounds become more basic. This occurs because of the electronegativity of the atoms as they move across periods. Nitrogen is less electronegative then oxygen‚ which means that the unshared pair of electrons of the conjugate base in an NH2 molecule (NH-) is in a higher energy orbital and therefore less stable than
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Ghiladi - Exam 1 Name: ______________‚ ______________ ID #:______________ (last) (first) Directions: 1. WRITE AND BUBBLE in LAST NAME‚ FIRST NAME and ID# on the Scantron sheet 2. Enter exam code 111 in the ‘Special Codes’ section of the Scantron sheet 3. Write your name and ID# on this exam paper 4. Hand in all materials before leaving the classroom 5. FAILURE TO FOLLOW DIRECTIONS WILL RESULT IN A ZERO h = 6.6262 x 10-34 J•s c = 3.00 x 108 m/s RH = 3.290 x 1015 s-1 Bohr radius
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